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Q.

Two cells of the same e.m.f. e but different internal resistances,R1 and R2, are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R [Fig.] is

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a

r1 - r2

b

r1 / r2

c

r1 r2

d

r1 + r2

answer is A.

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Detailed Solution

Current i in the circuit=2e/r1+r2+RVoltage drop across Ist cell=V1=e−ir1∴ V1=e−2er1R+r1+r2=0∴ e=2er1R+r1+r2 or R+r1+r2=2r1R=r1−r2

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